# Variance and Standard Deviation of multiple dice rolls

I'm trying to determine what the variance of rolling $5$ pairs of two dice are when the sums of all $5$ pairs are added up (i.e. ranging from $10$ to $60$).

My first question is, when I calculate the variance using $E[X^2]-E[X]^2$ I get $2.91$, but my Excel spreadsheet and other sites I've googled give $3.5$ with no explanation of what me taking place. Which one is correct?

Second, to calculate the variance of a random variable representing the sum of the $5$ pairs (i.e. between $10$ and $60$), is it simply $5 \times Var(X)$? What about the standard deviation, is it $\sigma \sqrt{n}$?

Last, is there any difference between calculating the dice sums as "$5$ pairs of $2$ dice" and "$10$ dice"? Will it make a practical difference? (I find it easier to calculate it as $10$ dice).

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It is not the variance, but the expected value of a dice roll, $E(X)$ that is 3.5. – wnvl Nov 14 '12 at 1:53
The assumptions in the second and third part of the question are all correct. – wnvl Nov 14 '12 at 1:55
So, in other words the standard deviation of 5 pairs of 2 dice and the standard deviation of 10 dice is 5.3759? Can you confirm? – Imray Nov 14 '12 at 2:02

So we are tossing $10$ dice. Let $X_i$ be the result of the $i$-th toss. Let $Y=X_1+X_2+\cdots +X_{10}$. It seems that you want the variance of $Y$.
The variance of a sum of independent random variables is the sum of the variances. Now calculate the variance of $X_i$. This as usual is $E(X_i^2)-(E(X_i))^2$.
We know that $E(X_i)=3.5$. For $E(X_i^2)$, note that this is $$\frac{1}{6}(1^2+2^2+3^2+4^2+5^2+6^2).$$